Small fixes to Appendix C
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@ -159,21 +159,21 @@ level 2/.style={sibling distance=0.7cm},
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\subsection{Proof of ~\Cref{lem:one-pass}}\label{sec:proof-one-pass}
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\paragraph{\onepass Correctness}
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We prove the correct computation of \prt, \lwght, \rwght values on \circuit by induction over the number of iterations in line ~\ref{alg:one-pass-loop} over the topological order \topord of the input circuit \circuit. Note that \topord is the standard definition of a topological ordering over the DAG structure of \circuit.
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We prove the correct computation of \prt, \lwght, \rwght values on \circuit by induction over the number of iterations in line~\ref{alg:one-pass-loop} over the topological order \topord of the input circuit \circuit. Note that \topord is the standard definition of a topological ordering over the DAG structure of \circuit.
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For the base case, we have only one gate, which by definition is a source gate and must be either \var or \tnum. In this case, as per \Cref{eq:T-all-ones}, lines ~\ref{alg:one-pass-var} and ~\ref{alg:one-pass-num} correctly compute \circuit.\prt as $1$ and \circuit.\val respectively.
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For the base case, we have only one gate, which by definition is a source gate and must be either \var or \tnum. In this case, as per \Cref{eq:T-all-ones}, lines~\ref{alg:one-pass-var} and~\ref{alg:one-pass-num} correctly compute \circuit.\prt as $1$ and \circuit.\val respectively.
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For the inductive hypothesis, assume that \onepass correctly computes \subcircuit.\prt, \subcircuit.\lwght, and \subcircuit.\rwght for all gates \gate in \circuit with $k > 0$ iterations over \topord.
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For the inductive hypothesis, assume that \onepass correctly computes \subcircuit.\prt, \subcircuit.\lwght, and \subcircuit.\rwght for all gates \gate in \circuit with $k \geq 0$ iterations over \topord.
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We now prove for $k + 1$ iterations that \onepass correctly computes the \prt, \lwght, and \rwght values for each gate $\gate_\vari{i}$ in \circuit. %By the hypothesis the first $k$ gates (alternatively \textit{iterations}) have correctly computed values.
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Note that the $\gate_\vari{k + 1}$ must be in the last ordering of all gates $\gate_\vari{i}$ for $i \in [k + 1]$. It is also the case that $\gate_{k+1}$ has two inputs. Finally, note that for \size(\circuit) > 1, if $\gate_{k+1}$ is a leaf node, we are back to the base case. Otherwise $\gate_{k + 1}$ is an internal node $\gate_\vari{s}.\type = \circplus$ or $\gate_\vari{s}.\type = \circmult$.
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We now prove for $k + 1$ iterations that \onepass correctly computes the \prt, \lwght, and \rwght values for each gate $\gate_\vari{i}$ in \circuit for $i \in [k + 1]$.
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Note that the $\gate_\vari{k + 1}$ must be in the last ordering of all gates $\gate_\vari{i}$. It is also the case that $\gate_{k+1}$ has two inputs. Finally, note that for \size(\circuit) > 1, if $\gate_{k+1}$ is a leaf node, we are back to the base case. Otherwise $\gate_{k + 1}$ is an internal node $\gate_\vari{s}.\type = \circplus$ or $\gate_\vari{s}.\type = \circmult$.
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When $\gate_{k+1}.\type = \circplus$, then by line ~\ref{alg:one-pass-plus} $\gate_{k+1}$.\prt $= \gate_{{k+1}_\lchild}$.\prt $+ \gate_{{k+1}_\rchild}$.\prt, a correct computation, as per \Cref{eq:T-all-ones}. Further, lines ~\ref{alg:one-pass-lwght} and ~\ref{alg:one-pass-rwght} compute $\gate_{{k+1}}.\lwght = \frac{\gate_{{k+1}_\lchild}.\prt}{\gate_{{k+1}}.\prt}$ and analogously for $\gate_{{k+1}}.\rwght$. Note that all values needed for each computation have been correctly computed by the I.H.
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When $\gate_{k+1}.\type = \circplus$, then by line ~\ref{alg:one-pass-plus} $\gate_{k+1}$.\prt $= \gate_{{k+1}_\lchild}$.\prt $+ \gate_{{k+1}_\rchild}$.\prt, a correct computation, as per \Cref{eq:T-all-ones}. Further, lines ~\ref{alg:one-pass-lwght} and ~\ref{alg:one-pass-rwght} compute $\gate_{{k+1}}.\lwght = \frac{\gate_{{k+1}_\lchild}.\prt}{\gate_{{k+1}}.\prt}$ and analogously for $\gate_{{k+1}}.\rwght$. Note that all values needed for each computation have been correctly computed by the inductive hypothesis.
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When $\gate_{k+1}.\type = \circmult$, then line ~\ref{alg:one-pass-mult} computes $\gate_{k+1}.\prt = \gate_{{k+1}_\lchild.\prt} \circmult \gate_{{k+1}_\rchild}.\prt$, which indeed is correct, as per \Cref{eq:T-all-ones}.
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\paragraph{\onepass Runtime}
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It is known that $\topord(G)$ is computable in linear time. Next, each of the $\size(\circuit)$ iterations of the loop in ~\Cref{alg:one-pass-loop} take $O\left( \multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit)}}\right)$ time. It is easy to see that all numbers that the algorithm computes is at most $\abs{\circuit}(1,\dots,1)$. Hence, by definition each such operation takes $\multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit)}}$ time, which proves the claimed runtime.
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It is known that $\topord(G)$ is computable in linear time. Next, each of the $\size(\circuit)$ iterations of the loop in \cref{alg:one-pass-loop} take $O\left( \multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit)}}\right)$ time. It is easy to see that each of all the numbers which the algorithm computes is at most $\abs{\circuit}(1,\dots,1)$. Hence, by definition each such operation takes $\multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit)}}$ time, which proves the claimed runtime.
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%In general it is known that an arithmetic computation which requires $M$ bits takes $O(\frac{\log{M}}{\log{N}})$ time for an input size $N$. Since each of the arithmetic operations at a given gate has a bit size of $O(\log{\abs{\circuit}(1,\ldots, 1)})$, thus, we obtain the general runtime of $O\left(\size(\circuit)\cdot \frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}}\right)$.
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@ -5,7 +5,7 @@
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\input{app_sample-monomial-pseudo-code}
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We briefly describe the top-down traversal of \sampmon. For a parent $+$ gate, the input to be visited is sampled from the weighted distribution precomputed by \onepass.
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When a parent $\times$ node is visited, both inputs are visited.
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The algorithm computes two properties: the set of all variable leaf nodes visited, and the product of signs of visited coefficient leaf nodes.
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The algorithm computes two properties: the set of all variable leaf nodes visited, and the product of the signs of visited coefficient leaf nodes.
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%
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We will assume the TreeSet data structure to maintain sets with logarithmic time insertion and linear time traversal of its elements.
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While we would like to take advantage of the space efficiency gained in using a circuit \circuit instead an expression tree \etree, we do not know that such a method exists when computing a sample of the input polynomial representation.
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@ -13,23 +13,25 @@ While we would like to take advantage of the space efficiency gained in using a
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The efficiency gains of circuits over trees is found in the capability of circuits to only require space for each \emph{distinct} term in the compressed representation. This saves space in such polynomials containing non-distinct terms multiplied or added to each other, e.g., $x^4$. However, to avoid biased sampling, it is imperative to sample from both inputs of a multiplication gate, independently, which is indeed the approach of \sampmon.
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\subsection{Proof of~\Cref{lem:sample}}\label{sec:proof-sample-monom}
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\begin{proof}
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%\paragraph*{\sampmon returns a valid monomial}
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We first need to show that $\sampmon$ indeed returns a monomial $\monom$,\footnote{Technically it returns $\var(\monom)$ but for less cumbersome notation we will refer to $\var(\monom)$ simply by $\monom$ in this proof.} such that $(\monom, \coef)$ is in $\expansion{\circuit}$, which we do by induction on the depth of $\circuit$.
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For the base case, let the depth $d$ of $\circuit$ be $0$. We have that the root node is either a constant $\coef$ for which by line ~\ref{alg:sample-num-return} we return $\{~\}$, or we have that $\circuit.\type = \var$ and $\circuit.\val = x$, and by line ~\ref{alg:sample-var-return} we return $\{x\}$. Both cases sample a monomial%satisfy ~\cref{def:monomial}
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For the base case, let the depth $d$ of $\circuit$ be $0$. We have that the root node is either a constant $\coef$ for which by line~\ref{alg:sample-num-return} we return $\{~\}$, or we have that $\circuit.\type = \var$ and $\circuit.\val = x$, and by line~\ref{alg:sample-var-return} we return $\{x\}$. Both cases sample a monomial%satisfy ~\cref{def:monomial}
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, and the base case is proven.
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For the inductive hypothesis, assume that for $d \leq k$ for some $k \geq 0$, that it is indeed the case that $\sampmon$ returns a monomial.
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For the inductive step, let us take a circuit $\circuit$ with $d = k + 1$. Note that each input has depth $d \leq k$, and by inductive hypothesis both of them return a valid monomial. Then the root can be either a $\circplus$ or $\circmult$ node. For the case of a $\circplus$ root node, line ~\ref{alg:sample-plus-bsamp} of $\sampmon$ will choose one of the inputs of the root. By inductive hypothesis it is the case that a monomial in \expansion(\circuit) is being returned from either input. Then it follows that for the case of $+$ root node a valid monomial is returned by $\sampmon$. When the root is a $\circmult$ node, line ~\ref{alg:sample-times-union} %and ~\ref{alg:sample-times-product} multiply
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For the inductive step, let us take a circuit $\circuit$ with $d = k + 1$. Note that each input has depth $d \leq k$, and by inductive hypothesis both of them return a valid monomial. Then the root can be either a $\circplus$ or $\circmult$ node. For the case of a $\circplus$ root node, line~\ref{alg:sample-plus-bsamp} of $\sampmon$ will choose one of the inputs of the root. By inductive hypothesis it is the case that a monomial in \expansion{\circuit} is being returned from either input. Then it follows that for the case of $+$ root node a valid monomial is returned by $\sampmon$. When the root is a $\circmult$ node, line~\ref{alg:sample-times-union} %and ~\ref{alg:sample-times-product} multiply
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computes the set union of the monomials returned by the two inputs of the root, and it is trivial to see
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%by definition ~\ref{def:monomial}
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%the product of two monomials is also a monomial, and
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by ~\Cref{def:expand-circuit} that \monom is a valid monomial in some $(\monom, \coef) \in \expansion{\circuit}$.
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by~\cref{def:expand-circuit} that \monom is a valid monomial in some $(\monom, \coef) \in \expansion{\circuit}$.
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We will next prove by induction on the depth $d$ of $\circuit$ that the $(\monom,\coef) \in \expansion{\circuit}$ is the \monom returned by $\sampmon$ with a probability %`that is in accordance with the monomial sampled,
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$\frac{|\coef|}{\abs{\circuit}\polyinput{1}{1}}$.
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For the base case $d = 0$, by definition ~\ref{def:express-tree} we know that the root has to be either a coefficient or a variable. For either case, the probability of the value returned is $1$ since there is only one value to sample from. When the root is a variable $x$ the algorithm correctly returns $(\{x\}, 1 )$. When the root is a coefficient, \sampmon ~correctly returns $(\{~\}, sign(\coef_i))$.
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For the base case $d = 0$, by definition~\ref{def:circuit} we know that the root has to be either a coefficient or a variable. For either case, the probability of the value returned is $1$ since there is only one value to sample from. When the root is a variable $x$ the algorithm correctly returns $(\{x\}, 1 )$. When the root is a coefficient, \sampmon ~correctly returns $(\{~\}, sign(\coef_i))$.
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For the inductive hypothesis, assume that for $d \leq k$ and $k \geq 0$ $\sampmon$ indeed samples $\monom$ in $(\monom, \coef)$ in $\expansion{\circuit}$ with probability $\frac{|\coef|}{\abs{\circuit}\polyinput{1}{1}}$.%bove is true.%lemma ~\ref{lem:sample} is true.
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@ -50,8 +52,8 @@ and we obtain the desired result.
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\paragraph{Run-time Analysis}
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It is easy to check that except for lines~\ref{alg:sample-times-union} and~\ref{alg:sample-plus-bsamp}, all lines take $O(1)$ time. For \Cref{alg:sample-times-uinon}, consider an execution of~\Cref{alg:sample-times-union}. We note that we will be adding a given set of variables to some set at most once: since the sum of the sizes of the sets at a given level is at most $\degree(\circuit)$, each gate visited takes $O(\log{\degree(\circuit)})$. For \Cref{alg:sample-plus-bsamp}, note that we pick $\circuit_\linput$ with probability $\frac a{a+b}$ where $a=\circuit.\vari{Lweight}$ and $b=\circuit.\vari{Rweight}$. We can implement this step by picking a random number $r\in[a+b]$ and then checking if $r\le a$. It is easy to check that $a+b\le \abs{\circuit}(1,\dots,1)$. This means we need to add and compare $\log{\abs{\circuit}(1,\ldots, 1)}$-bit numbers, which can certainly be done in time $\multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit)}}$ (note that this is an over-estimate).
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%\paragraph*{Run-time Analysis}
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It is easy to check that except for lines~\ref{alg:sample-times-union} and~\ref{alg:sample-plus-bsamp}, all lines take $O(1)$ time. For \Cref{alg:sample-times-union}, consider an execution of~\Cref{alg:sample-times-union}. We note that we will be adding a given set of variables to some set at most once: since the sum of the sizes of the sets at a given level is at most $\degree(\circuit)$, each gate visited takes $O(\log{\degree(\circuit)})$. For \Cref{alg:sample-plus-bsamp}, note that we pick $\circuit_\linput$ with probability $\frac a{a+b}$ where $a=\circuit.\vari{Lweight}$ and $b=\circuit.\vari{Rweight}$. We can implement this step by picking a random number $r\in[a+b]$ and then checking if $r\le a$. It is easy to check that $a+b\le \abs{\circuit}(1,\dots,1)$. This means we need to add and compare $\log{\abs{\circuit}(1,\ldots, 1)}$-bit numbers, which can certainly be done in time $\multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit)}}$ (note that this is an over-estimate).
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% we have $> O(1)$ time when $\abs{\circuit}(1,\ldots, 1) > \size(\circuit)$. when this is the case that for each sample, we have $\frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\log{\size(\circuit)}}$ operations, since we need to read in and then compare numbers of of $\log{{\abs{\circuit}(1,\ldots, 1)}}$ bits.
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Denote \cost(\circuit) (\Cref{eq:cost-sampmon}) to be an upper bound of the number of nodes visited by \sampmon. Then the runtime is $O\left(\cost(\circuit)\cdot \log{\degree(\circuit)}\cdot \multc{\log\left(\abs{\circuit(1\ldots, 1)}\right)}{\log{\size(\circuit)}}\right)$.
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@ -139,4 +141,7 @@ As in the $\circmult$ case the \emph{reduced} invariant of \reduce implies that
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Similar to the case of $\circuit.\type = \circmult$, (\ref{eq:plus-rhs}) follows by equations $(\ref{eq:cost-sampmon})$ and $(\ref{eq:ih-bound-cost})$.
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This proves (\ref{eq:strict-upper-bound}) for the $\circplus$ case, as desired. % and thus the claimed $O(k\log{k}\cdot \frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\size(\circuit)}\cdot\depth(\circuit))$ runtime for $k = \degree(\circuit)$ follows.
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This proves (\ref{eq:strict-upper-bound}) for the $\circplus$ case, as desired.
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\end{proof}
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% and thus the claimed $O(k\log{k}\cdot \frac{\log{\abs{\circuit}(1,\ldots, 1)}}{\size(\circuit)}\cdot\depth(\circuit))$ runtime for $k = \degree(\circuit)$ follows.
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@ -3,7 +3,7 @@
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\caption{\sampmon(\circuit)}
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\label{alg:sample}
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\begin{algorithmic}[1]
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\revision{\Require \circuit: Circuit}
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\Require \circuit: Circuit
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\Ensure \vari{vars}: TreeSet
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\Ensure \vari{sgn} $\in \{-1, 1\}$
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\Comment{\Cref{alg:one-pass-iter} should have been run before this one} % algorithm ~\ref{alg:sample}}
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@ -114,7 +114,7 @@ The algorithm (\approxq detailed in \Cref{alg:mon-sam}) to prove~\Cref{lem:appro
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\label{eq:tilde-Q-bi}
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\rpoly\inparen{X_1,\dots,X_\numvar}=\hspace*{-1mm}\sum_{(\monom,\coef)\in \expansion{\circuit}} \hspace*{-2mm} \indicator{\monom\mod{\mathcal{B}}\not\equiv 0}\cdot \coef\cdot\hspace*{-2mm}\prod_{X_i\in \var\inparen{\monom}}\hspace*{-2mm} X_i
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\end{equation}
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Given the above, the algorithm is a sampling based algorithm for the above sum: we sample (via \sampmon) $(\monom,\coef)\in \expansion{\circuit}$ with probability proportional%\footnote{We could have also uniformly sampled from $\expansion{\circuit}$ but this gives better parameters.}
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Given the above, the algorithm is a sampling based algorithm for the above sum: we sample (via \sampmon) $(\monom,\coef)\in \expansion{\circuit}$ with probability proportional %\footnote{We could have also uniformly sampled from $\expansion{\circuit}$ but this gives better parameters.}
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to $\abs{\coef}$ and compute $Y=\indicator{\monom\mod{\mathcal{B}}\not\equiv 0}\cdot \prod_{X_i\in \var\inparen{\monom}} p_i$. Taking $\numsamp$ samples and computing the average of $Y$ gives us our final estimate. \onepass is used to compute the sampling probabilities needed in \sampmon (details are in~\Cref{sec:proofs-approx-alg}).
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%\approxq (\cref{alg:mon-sam}) modifies \circuit with a call to \onepass. It then samples from $\circuit_{\vari{mod}}\numsamp$ times and uses that information to approximate $\rpoly$.
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@ -76,7 +76,7 @@ This follows from~\Cref{lem:circuits-model-runtime} (\cref{sec:circuit-runtime})
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%
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We make a simple observation to conclude the presentation of our results.
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So far we have only focused on the expectation of $\poly$. In addition, we could e.g. prove bounds of probability of the multiplicity being at least $1$. Progress can be made on this as follows:
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For any positive integer $m$ we can compute the $m$-th moment of the multiplicities, allowing us to e.g. to use Chebyschev inequality or other high moment based probability bounds on the events we might be interested in.
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For any positive integer $m$ we can compute the $m$-th moment of the multiplicities, allowing us to e.g. use Chebyschev inequality or other high moment based probability bounds on the events we might be interested in.
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We leave a further investigation of this question for future work.
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%%% Local Variables:
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@ -88,7 +88,7 @@
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\newcommand{\lchild}{\vari{L}}
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\newcommand{\rchild}{\vari{R}}
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%members of T
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\newcommand{\val}{\vari{val}\xspace}
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\newcommand{\val}{\vari{val}}
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\newcommand{\wght}{\vari{weight}\xspace}
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\newcommand{\vpartial}{\vari{partial}\xspace}
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%types of T
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@ -117,7 +117,7 @@
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\newcommand{\degree}{\func{deg}}
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\newcommand{\size}{\func{size}}
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\newcommand{\depth}{\func{depth}}
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\newcommand{\topord}{\func{TopOrd}}
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\newcommand{\topord}{\func{TopOrd}\xspace}
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%saving \treesize for now to keep latex from breaking
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\newcommand{\treesize}{\func{size}}
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\newcommand{\sign}{\func{sgn}}
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